What is the general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi ?

The general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi is x= pi/2

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3sin^2(x)+sin(x)-4=0,0<= x<= 2pi

3 sin^{2} (x) + sin (x) - 4 = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}

Solution steps

3 sin^{2} (x) + sin (x) - 4 = 0, 0 \leq x \leq 2 π

Solve by substitution

sin (x) = 1, sin (x) = - \frac{4}{3}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

sin (x) = - \frac{4}{3}, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{π}{2}

sin (θ) + 2 = 0

2 tan (2 x) - 10 = 3.4641

8 cos^{2} (x) + 14 cos (x) = - 3

4 sin (x) cos (x) - 23 \cdot sin (x) - 2 cos (x) + 3 = 0

2 sin^{2} (θ) - 2 = 3 sin (θ)

What is the general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi ?
The general solution for 3sin^2(x)+sin(x)-4=0,0<= x<= 2pi is x= pi/2